To solve the system of equations 3x^2-5y^2+2y=453x2−5y2+2y=45 and y=2x+10y=2x+10, just substitute value of yy from latter equation in the former equation.
i.e. 3x^2-5(2x+10)^2+2(2x+10)=453x2−5(2x+10)2+2(2x+10)=45
or 3x^2-5(4x^2+40x+100)+4x+20=453x2−5(4x2+40x+100)+4x+20=45
or -17x^2-196x-525=0−17x2−196x−525=0 or 17x^2+196x+525=017x2+196x+525=0
Hence x=(-196+-sqrt(196^2-4*17*525))/34x=−196±√1962−4⋅17⋅52534
= (-196+-sqrt(38416-35700))/34−196±√38416−3570034
= (-196+-sqrt2716)/34=(-196+-52.12)/34−196±√271634=−196±52.1234
i.e. x=-4.2x=−4.2 or -7.3−7.3
and y=2xx(-4.2)+10=1.6y=2×(−4.2)+10=1.6 or y=2xx(-7.3)+10=-4.6y=2×(−7.3)+10=−4.6
and Solution set is (-4.2,1.6)(−4.2,1.6) and (-7.3,-4.6)(−7.3,−4.6)
This is intersection of a hyperbola and line.
graph{(3x^2-5y^2+2y-45)(2x+10-y)=0 [-20, 20, -10, 10]}
